hdu4753 狀態壓縮dp博弈（記憶化搜索寫法）

 

Problem Description 　　　There is a 3 by 3 grid and each vertex is assigned a number.


　　　It looks like JiuGongGe, but they are different, for we are not going to fill the cell but the edge. For instance,


adding edge 6 –> 10
　　　The rule of this game is that each player takes turns to add an edge. You will get one point if the edge you just added, together with edges already added before, forms a new square (only square of size 1 is considered). Of course, you get two points if that edge forms two squares. Notice that an edge can be added only once.


forming two squares to get two points
　　Tom200 and Jerry404 is playing this little game, and have played n rounds when Fishhead comes in. Fishhead wants to know who will be the winner. Can you help him? Assume that Tom200 and Jerry404 are clever enough to make optimal decisions in each round. Every Game starts from Tom200.

Input 　　The first line of the input contains a single integer T (T <= 100), the number of test cases.
　　For each case, the first line contains an integers n (12 <= n <= 24), which means they have taken total n rounds in turn. Next n lines each contains two integers a, b (a, b <= 16) representing the two endpoints of the edge.

Output 　　For each case output one line Case #X: , representing the Xth case, starting from 1. If Tom200 wins, print Tom200 on one line, print Jerry404 otherwise.

Sample Input

1
15
1 2
1 5
2 6
5 9
6 10
9 10
5 6
2 3
3 7
7 11
10 11
3 4
6 7
7 8
4 8

Sample Output

Case #1: Tom200
Hint
　　In case 1, Tom200 gets two points when she add edge 5 -> 6, two points in edge 6 -> 7, one point in 4 -> 8.

/**
hdu4753 狀態壓縮dp博弈（記憶化搜索寫法）
題目大意:為一個3*3的棋盤連邊（如圖所示）加上當前邊後多了幾個格子四條邊都被填上了，就得幾分，現在tom和jerry交替填邊，總是tom先，
          現在已經填了n條邊了，問剩下的邊二者都采取最優的策略，誰會贏
解題思路：由於未知的邊最多只有12條，我們可以采取狀態壓縮，用搜索的方式寫
注意：不可以直接壓24位，會爆內存的
*/
#include 
#include 
#include 
#include 
using namespace std;
int mp[24][24];///給每條邊進行編號
int vis[25];///標記是否已經訪問過該邊
int dp[(1<<13)+3],cnt,x[25];
int tom,jerry,n;
int circle[9][4]={///對應的9個格子各自的四條邊
1,4,5,8,
2,5,6,9,
3,6,7,10,
8,11,12,15,
9,12,13,16,
10,13,14,17,
15,18,19,22,
16,19,20,23,
17,20,21,24
};
void init()
{
    memset(mp,0,sizeof(mp));
    mp[1][2]=1,mp[2][3]=2,mp[3][4]=3;
    mp[1][5]=4,mp[2][6]=5,mp[3][7]=6,mp[4][8]=7;
    mp[5][6]=8,mp[6][7]=9,mp[7][8]=10;
    mp[5][9]=11,mp[6][10]=12,mp[7][11]=13,mp[8][12]=14;
    mp[9][10]=15,mp[10][11]=16,mp[11][12]=17;
    mp[9][13]=18,mp[10][14]=19,mp[11][15]=20,mp[12][16]=21;
    mp[13][14]=22,mp[14][15]=23,mp[15][16]=24;
}
int judge()
{
    int sum=0;
    for(int i=0;i<9;i++)
    {
        if(vis[circle[i][0]]&&vis[circle[i][1]]&&vis[circle[i][2]]&&vis[circle[i][3]])
             sum++;
    }
    return sum;
}

int ok(int ste,int k)
{
    int vv[25];
    memset(vv,0,sizeof(vv));
    for(int i=0;i<=cnt;i++)
    {
        if(ste&(1<maxx)
                maxx=k-sum;
        }
    }
    return dp[ste]=maxx;
}
int main()
{
    int T,tt=0;
    scanf(%d,&T);
    init();
    while(T--)
    {
        memset(vis,0,sizeof(vis));
        scanf(%d,&n);
        tom=0,jerry=0;
        int s=0,ste=0;
        for(int i=0;iy)swap(x,y);
            vis[mp[x][y]]=1;
            int s1=judge();
            if(i&1)
            {
                jerry+=(s1-s);
            }
            else
            {
                tom+=(s1-s);
            }
            s=s1;
           /// printf((%d %d)
,tom,jerry);
        }
        cnt=-1;
        for(int i=1;i<=24;i++)
        {
            if(!vis[i]) x[++cnt]=i;
        }
        s=9-judge();
        memset(dp,-1,sizeof(dp));
        int num=dfs(0,s);
        printf(Case #%d: ,++tt);
        if((n&1)==0)
        {
            if(tom+num>jerry+s-num)
                printf(Tom200
);
            else
                printf(Jerry404
);
        }
        else
        {
            if(tom+s-num>jerry+num)
                printf(Tom200
);
            else
                printf(Jerry404
);
        }
    }
    return 0;
}